What is the extraneous solution to these equations? $\dfrac{x^2 + 33}{x + 7} = \dfrac{x + 89}{x + 7}$
Explanation: Multiply both sides by $x + 7$ $ \dfrac{x^2 + 33}{x + 7} (x + 7) = \dfrac{x + 89}{x + 7} (x + 7)$ $ x^2 + 33 = x + 89$ Subtract $x + 89$ from both sides: $ x^2 + 33 - (x + 89) = x + 89 - (x + 89)$ $ x^2 + 33 - x - 89 = 0$ $ x^2 - 56 - x = 0$ Factor the expression: $ (x + 7)(x - 8) = 0$ Therefore $x = -7$ or $x = 8$ At $x = -7$ , the denominator of the original expression is 0. Since the expression is undefined at $x = -7$, it is an extraneous solution.